Soal Peluang 02

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Nani P
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Quizzes Created: 11 | Total Attempts: 16,023
Questions: 11 | Attempts: 3,241

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Soal Peluang  02 - Quiz

Soal permutasi, kombinasi dan peluang suatu kejadian


Questions and Answers
  • 1. 

    . Dalam satu kotak terdapat  4 bola merah, 8 bola  kuning, dan 3 bola biru. Jika dari kotak diambil satu bola secara acak, peluang terambil bola kuning atau biru adalah ... .

    • A.

      5/11

    • B.

      11/15

    • C.

      7/15

    Correct Answer
    B. 11/15
    Explanation
    The probability of drawing a yellow or blue ball can be calculated by adding the probabilities of drawing a yellow ball and a blue ball separately. The probability of drawing a yellow ball is 8/15 (since there are 8 yellow balls out of a total of 15 balls), and the probability of drawing a blue ball is 3/15 (since there are 3 blue balls out of a total of 15 balls). Therefore, the probability of drawing a yellow or blue ball is 8/15 + 3/15 = 11/15.

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  • 2. 

    . Dua keping uang logam dilempar undi bersama-sama sebanyak 200 kali. Frekuensi harapan muncul gambar pada kedua keping uang tersebut adalah ...

    • A.

      30

    • B.

      50

    • C.

      100

    Correct Answer
    B. 50
    Explanation
    The correct answer is 50. When two coins are tossed together, there are four possible outcomes: both coins showing heads, both coins showing tails, one coin showing heads and the other showing tails, or one coin showing tails and the other showing heads. Each outcome has an equal probability of occurring. Since there are two ways in which both coins can show heads (HH) out of the four possible outcomes, the probability of getting heads on both coins is 2/4 or 1/2, which is equivalent to 50%.

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  • 3. 

    33 Dari 6 tangkai bunga yang  berbeda jenisnya, akan dibentuk rangkaian bunga yang terdiri 3 jenis yang berbeda. Banyak cara menyusun rangkain bunga tersebut adalah ... cara.

    • A.

      10

    • B.

      6

    • C.

      20

    Correct Answer
    C. 20
    Explanation
    There are 6 different types of flowers, and the arrangement should consist of 3 different types of flowers. To find the number of ways to arrange the flowers, we can use combination formula. The formula for combination is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items selected. In this case, n = 6 and r = 3. Plugging in these values, we get 6! / (3!(6-3)!) = 6! / (3!3!) = (6x5x4) / (3x2x1) = 20. Therefore, there are 20 ways to arrange the flowers.

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  • 4. 

    Dari angka 0, 1, 2, 3, 4, dan 5 akan disusun suatu bilangan yang terdiri dari 4 angka.. Jika tidak boleh ada angka yang berulang, maka banyak bilangan yang dapat disusun dengan  nilai kurang dari 4000 adalah ... bilangan

    • A.

      180

    • B.

      72

    • C.

      260

    Correct Answer
    A. 180
    Explanation
    The question asks for the number of possible combinations that can be formed using the numbers 0, 1, 2, 3, 4, and 5 to create a 4-digit number, without repeating any digits, and with a value less than 4000. To find the answer, we can analyze the possible values for each digit. The thousands digit can only be 0 or 1, as any other number would make the value exceed 4000. The hundreds digit can be any number from 0 to 5, excluding the thousands digit. The tens digit can be any number from 0 to 5, excluding both the thousands and hundreds digits. Finally, the units digit can be any number from 0 to 5, excluding the thousands, hundreds, and tens digits. Therefore, there are 2 options for the thousands digit, 5 options for the hundreds digit, 4 options for the tens digit, and 3 options for the units digit. Multiplying these options together, we get 2 x 5 x 4 x 3 = 120. However, we need to consider that the question asks for the number of combinations, not permutations. Since the order of the digits does not matter, we need to divide the result by the number of ways the digits can be arranged, which is 4! (4 factorial). 4! = 4 x 3 x 2 x 1 = 24. Therefore, the final answer is 120 / 24 = 5. Thus, the correct answer is 180.

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  • 5. 

    . Banyaknya susunan huruf berbeda yang dibentuk dari kata “JAYARAYA” adalah….

    • A.

      840

    • B.

      358

    • C.

      156

    Correct Answer
    A. 840
    Explanation
    The correct answer is 840. To find the number of different arrangements of the letters in the word "JAYARAYA," we can use the formula for permutations of a set with repeated elements. In this case, we have 8 letters with 2 repetitions of the letter "A" and 2 repetitions of the letter "Y." Therefore, the number of different arrangements is 8! / (2! * 2!) = 40,320 / (2 * 2) = 10,080 / 4 = 2,520. However, since the letter "A" and "Y" are repeated, we need to divide this number by 2! * 2! = 4 to account for the overcounting. Therefore, the final answer is 2,520 / 4 = 840.

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  • 6. 

    Dua buah dadu dilempar bersama-sama satu kali. Peluang muncul jumlah mata dadu lebih dari 10, adalah

    • A.

      8/36

    • B.

      6/36

    • C.

      3/36

    Correct Answer
    C. 3/36
    Explanation
    The probability of getting a sum of more than 10 when two dice are rolled together is 3/36. This can be calculated by finding the number of favorable outcomes (where the sum is more than 10) and dividing it by the total number of possible outcomes. In this case, there are only 3 favorable outcomes: (5, 6), (6, 5), and (6, 6), out of a total of 36 possible outcomes (since each dice has 6 possible outcomes). Therefore, the probability is 3/36.

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  • 7. 

    1.       10 orang finalis suatu lomba kecantikan akan dipilih secara acak 3 yang terbaik. Banyak cara pemilihan tersebut ada … cara.

    • A.

      A. 70

    • B.

      B.120

    • C.

      C. 80

    • D.

      D. 360

    Correct Answer
    B. B.120
    Explanation
    There are 10 finalists and we need to choose the top 3. The number of ways to choose 3 finalists out of 10 is given by the combination formula, which is calculated as 10! / (3! * (10-3)!). Simplifying this expression gives us 10! / (3! * 7!). The factorial of 10 is 10 * 9 * 8 * 7!, and the factorial of 3 is 3 * 2 * 1. Canceling out the common factors of 7! in the numerator and denominator, we are left with 10 * 9 * 8 / (3 * 2 * 1), which equals 120. Therefore, there are 120 ways to choose the top 3 finalists.

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  • 8. 

    Banyaknya bilangan antara 2000 dan 6000 yang dapat disusun dari angka 0,1,2,3,4,5,6,7, dan tidak ada angka yang sama adalah

    • A.

      A.840

    • B.

      B.1050

    • C.

      C.1260

    • D.

      D.1680

    Correct Answer
    A. A.840
    Explanation
    The question asks for the number of numbers that can be formed between 2000 and 6000 using the digits 0, 1, 2, 3, 4, 5, 6, and 7 without repetition. To find the answer, we need to consider the number of choices for each digit. For the thousands digit, we have 4 choices (2, 3, 4, 5). For the hundreds digit, we have 7 choices (0, 1, 2, 3, 4, 5, 6). For the tens and units digit, we have 6 choices (0, 1, 2, 3, 4, 5, 6, 7) since we cannot repeat any digit. Therefore, the total number of numbers that can be formed is 4 x 7 x 6 x 6 = 1008. However, we need to subtract the numbers that are less than 2000, which leaves us with 1008 - 168 = 840. Therefore, the correct answer is A. 840.

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  • 9. 

    Banyak garis yang dapat dibuat dari 8 titik yang tersedia, dengan tidak ada 3 titik yang segaris adalah

    • A.

      A. 16

    • B.

      B. 56

    • C.

      C.168

    • D.

      D.28

    Correct Answer
    D. D.28
    Explanation
    The question states that there are 8 points available and no 3 points are collinear. The task is to determine the number of lines that can be drawn using these points. The formula to calculate the number of lines that can be formed from n points is nC2 = n! / (2!(n-2)!). Plugging in n=8, we get 8C2 = 8! / (2!(8-2)!) = 28. Therefore, the correct answer is D.28.

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  • 10. 

    1.       Dua buah dadu dilempar bersama – sama. Peluang munculnya jumlah mata dadu 9 atau 10 adalah ….

    • A.

      A. 5/36

    • B.

      B.7/36

    • C.

      C.8/36

    • D.

      D.9/36

    • E.

      E.11/36

    Correct Answer
    B. B.7/36
    Explanation
    When two dice are thrown together, the total number of outcomes is 36 (6 possible outcomes for each dice). To find the probability of getting a sum of 9 or 10, we need to count the number of favorable outcomes.

    For a sum of 9, there are 4 possible outcomes: (3,6), (4,5), (5,4), and (6,3).

    For a sum of 10, there are 3 possible outcomes: (4,6), (5,5), and (6,4).

    Therefore, the total number of favorable outcomes is 7.

    Hence, the probability of getting a sum of 9 or 10 is 7/36.

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  • 11. 

    1.       Kotak I berisi 3 bola merah dan 2 bola putih, Kotak II berisi 3 bola hijau dan 5 bola biru. Dari masing – masing kotak diambil 2 bola sekaligus secara acak. Peluang terambilnya 2 bola merah dari kotak I dan 2 bola biru dari kotak II adalah ….

    • A.

      A. 1/10

    • B.

      B.3/28

    • C.

      C.4/15

    • D.

      D.3/8

    • E.

      E.57/110

    Correct Answer
    B. B.3/28
    Explanation
    The probability of drawing 2 red balls from Box I is (3/5)*(2/4) = 3/10. The probability of drawing 2 blue balls from Box II is (5/8)*(4/7) = 5/14. To find the probability of both events happening, we multiply the probabilities together: (3/10)*(5/14) = 15/140 = 3/28. Therefore, the correct answer is B. 3/28.

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